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Question Number 79756 by mathmax by abdo last updated on 27/Jan/20

$$calculate{lim}_{x\to 1}\frac{sin\left(\pi x\right)}{1-x^2}withouthospitalrule$$

Commented by john santu last updated on 27/Jan/20

$$\underset{x\to 1}{\lim }\left(\frac{\sin (\pi -\pi \mathrm{x})}{(\pi -\pi \mathrm{x})}\right)\times \underset{x\to 1}{\lim }\frac{(\pi -\pi \mathrm{x})}{1-{\mathrm{x}}^2}$$$$=1\times \underset{x\to 1}{\lim }\frac{\pi (1-\mathrm{x})}{(1-\mathrm{x})(1+\mathrm{x})}$$$$=\frac{\pi }{2}$$

Commented by mathmax by abdo last updated on 28/Jan/20

$$changement1-x=tgive$$$${lim}_{x\to 1}\frac{sin\left(\pi x\right)}{1-x^2}={lim}_{t\to 0}\frac{sin\left(\pi (1-t)\right)}{1-{(1-t)}^2}$$$$={lim}_{t\to 0}\frac{sin\left(\pi t\right)}{1-(t^2-2t+1)}={lim}_{t\to 0}\frac{sin\left(\pi t\right)}{2t-t^2}$$$$={lim}_{t\to 0}\frac{sin\left(\pi t\right)}{\pi t}\times \frac{\pi t}{2t-t^2}={lim}_{t\to 0}\frac{sin\left(\pi t\right)}{\pi t}\times {lim}_{t\to 0}\frac{\pi }{2-t}$$$$=1\times \frac{\pi }{2}=\frac{\pi }{2}$$

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